相关疑难解决方法(0)

如何将Java赋值表达式转换为Kotlin

像java这样的东西

int a = 1, b = 2, c = 1;
if ((a = b) !=c){
    System.out.print(true);
}
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现在它应该转换为kotlin之类的

var a:Int? = 1
var b:Int? = 2
var c:Int? = 1
if ( (a = b) != c)
    print(true)
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但这不正确.

这是我得到的错误:

in " (a=b)" Error:(99, 9) Kotlin: Assignments are not expressions, and only expressions are allowed in this context
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实际上,上面的代码只是澄清问题的一个例子.这是我的原始代码:

fun readFile(path: String): Unit { 
    var input: InputStream = FileInputStream(path) 
    var string: String = "" 
    var tmp: Int = -1 …
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java kotlin

30
推荐指数
4
解决办法
7460
查看次数

表达时不允许分配?

在Java中,我们通常可以在while条件内执行赋值.然而Kotlin抱怨它.所以下面的代码不能编译:

val br = BufferedReader(InputStreamReader(
        conn.inputStream))

var output: String
println("Output from Server .... \n")
while ((output = br.readLine()) != null) { // <--- error here: Assignments are not expressions, and only expressions are allowed in this context
    println(output)
}
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根据这个其他线程,这似乎是最好的解决方案:

val reader = BufferedReader(reader)
var line: String? = null;
while ({ line = reader.readLine(); line }() != null) { // <--- The IDE asks me to replace this line for while(true), what the...?
  System.out.println(line);
}
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但是吗?

kotlin kotlin-interop

17
推荐指数
3
解决办法
5305
查看次数

标签 统计

kotlin ×2

java ×1

kotlin-interop ×1